AMC 10 · 2019 · #16

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Nested Frame / Concentric Difference

area-circlesequilateral-triangletangent-circlesthirty-sixty-ninety-trianglearea-difference area-differenceidentify-subproblems ↑ Prerequisites: area-circlesequilateral-triangletangent-circles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Thirteen unit circles (radius 1) sit inside one larger circle. Every touching point is a tangency (no real overlap). The shaded region is the part of the big disk that is outside all 13 small disks. Find that shaded area.

Pick an answer.

(A)

$4 \pi \sqrt{3}$

(B)

$7 \pi$

(C)

$\pi\left(3\sqrt{3} +2\right)$

(D)

$10 \pi \left(\sqrt{3} - 1\right)$

(E)

$\pi\left(\sqrt{3} + 6\right)$

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems): the shaded area is just (big circle area) - (13 × small circle area). So we only need the big circle's radius R. Tool #1 (Diagram): mark the centers of the small circles. Three outer small-circle centers form an equilateral triangle of side 2 around the origin, which pins R via simple right-triangle geometry.

1STEP 1

Tangent unit circles touch at one point, so two neighboring centers lie exactly 2 apart — pinning the first inner ring.

neighbor centers distance = 1+1 = 2

2STEP 2

By the figure the top outermost center sits at (0, 2√(3)), so every outer-ring center lies 2√(3) from the origin.

|outer-ring center| = 2√(3)

3STEP 3

That outer-ring circle is internally tangent to the big one, so the big radius is R = 2√(3) + 1.

R = 2√(3) + 1

4STEP 4

Square it: (2√(3)+1)² = 12 + 4√(3) + 1, so the big circle's area is π(13 + 4√(3)).

π R² = π(13 + 4√(3))

5STEP 5

Each unit circle has area π, and 13 of them give a combined 13π.

13 · π = 13π

6STEP 6

Subtract the 13π: π(13 + 4√(3)) - 13π = 4π√(3), matching choice (A).

π(13 + 4√(3)) - 13π = 4π√(3) → (A)

Answer

4 π √(3)

4π√(3) ≈ 4 · 3.14 · 1.73 ≈ 21.8. The big disk has area π(13 + 4√(3)) ≈ π · 19.93 ≈ 62.6 and the 13 small disks together are 13π ≈ 40.8. Difference ≈ 21.8 — matches. Also choices (B) 7π ≈ 22.0 is dangerously close numerically, but it would only appear if you mis-squared (2√(3)+1)² as 12 + 1 = 13, forgetting the 4√(3) cross term — exactly the trap.

💡Key takeaway

This AMC 10 problem only needs Grade 8 right-triangle geometry you already know — once you see the big radius is 2√(3)+1, the area π(13+4√(3)) - 13π collapses to just 4π√(3). The answer is (A).

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